A) 2.7 kcal
B) -2.7 kcal
C) 9.3 kcal
D) -9.3 kcal
Correct Answer: B
Solution :
The change in Gibbs free energy is given by \[\Delta G=\Delta H-T\Delta S\] where, \[\Delta H\] = enthalpy of the reaction \[\Delta S\] = entropy of the reaction Thus, in order to determine\[\Delta G\], the values of \[\Delta H\]must be known, the value of \[\Delta H\]can be calculated by the equation \[\Delta H=\Delta U+\Delta {{n}_{g}}RT\] where \[\Delta U\] = change in internal energy \[\Delta {{n}_{g}}\] = (number of moles of gaseous products) - (number of moles of gaseous reactants) = 2 - 0 = 2 R = gas constant = 2 cal But, \[\Delta H=\Delta u+\Delta {{n}_{g}}RT\] \[\Delta u=2.1kcal=2.1\times {{10}^{3}}cal\] \[[\because \,1kcal\,={{10}^{3}}cal]\] \[\therefore \] \[\Delta H=(2.1\times {{10}^{3}})+(2\times 2\times 300)=3300cal\] Hence, \[\Delta G=\Delta H-T\Delta S\] Þ \[\Delta G=(3300)-(300-20)\] \[\Delta G=-2700\,cal\] \[\therefore \] \[\Delta G=-2.7kcal\]
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